Question

Prove that for any prime positive interger $p,\sqrt{p}$ is an irrational number

Answer 1

SOLUTION :  

Let us assume that √p is rational. Then,it will be of the form a/b where a, b are co primes integers and b ≠0.

Let a and b have no common factor other than 1.

√p = a/b

on squaring both sides,

(√p)² = (a/b)²

p = a²/b²

pb² = a² ……….(1)

p divides a²  

p divide a ……….(2)

[By Theorem , If p divides a², then p divide a]

Let a = pc  

On squaring both sides  

a² = p²c²

Put the value of a² in eq 1,  

pb² = a²

pb² = p²c²

b² = p²c²/p

b² = pc²

p divides b²

p divide b …………(3)

[By Theorem , If p divides a², then p divide a]

From eq 2 p divide a and from eq 3 p divide b. It means p is a factor of both a and b.

This contradicts the fact that a and b are co primes. So ,our assumption is wrong.

Hence√ p is irrational.

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